Problem: Complete the square to solve for $x$. $4x^{2}-28x+49 = 0$
First, divide the polynomial by $4$ , the coefficient of the $x^2$ term. $x^2 - 7x + \dfrac{49}{4} = 0$ Note that the left side of the equation is already a perfect square trinomial. The coefficient of our $x$ term is $-7$ , half of it is $-\dfrac{7}{2}$ , and squaring it gives us ${\dfrac{49}{4}}$ , our constant term. Thus, we can rewrite the left side of the equation as a squared term. $( x - \dfrac{7}{2} )^2 = 0$ Take the square root of both sides. $x - \dfrac{7}{2} = 0$ Isolate $x$ to find the solution(s). The solution is: $x = \dfrac{7}{2}$ We already found the completed square: $( x - \dfrac{7}{2} )^2 = 0$